If [tex]$32 A+3 B=\left[\begin{array}{ll}2 & 3 \\ 4 & 0\end{array}\right]$[/tex] and [tex]$3 A+2 B=\left[\begin{array}{cc}-2 & 2 \\ 1 & -5\end{array}\right]$[/tex], find [tex]$A$[/tex] and [tex]$B$[/tex].
Answer (2)
The matrices A and B are found by first transforming the given equations to eliminate one matrix, solving for A, and then substituting to find B. The final results are: A = \begin{bmatrix} \frac{2}{11} & 0 \ \frac{1}{11} & \frac{3}{11} \end{bmatrix} and B = \begin{bmatrix} -\frac{14}{11} & 1 \ \frac{4}{11} & -\frac{32}{11} \end{bmatrix}.
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Multiply the first equation by 2 and the second equation by 3 to prepare for eliminating B.
Subtract the modified equations to eliminate B and solve for A: A = 55 1 [ 10 5 0 15 ] = [ 11 2 11 1 0 11 3 ]
Substitute the value of A into one of the original equations and solve for B: B = 2 1 [ − 11 28 11 8 11 22 − 11 64 ] = [ − 11 14 11 4 1 − 11 32 ]
The solution is: A = [ 11 2 11 1 0 11 3 ] , B = [ − 11 14 11 4 1 − 11 32 ]
Explanation
Problem Setup We are given two matrix equations:
32 A + 3 B = [ 2 4 3 0 ]
3 A + 2 B = [ − 2 1 2 − 5 ]
Our goal is to find the matrices A and B .
Eliminating B Let's multiply the first equation by 2 and the second equation by 3. This will allow us to eliminate B when we subtract the equations:
2 ( 32 A + 3 B ) = 2 [ 2 4 3 0 ] ⇒ 64 A + 6 B = [ 4 8 6 0 ]
3 ( 3 A + 2 B ) = 3 [ − 2 1 2 − 5 ] ⇒ 9 A + 6 B = [ − 6 3 6 − 15 ]
Solving for A Now, subtract the second equation from the first to eliminate B :
( 64 A + 6 B ) − ( 9 A + 6 B ) = [ 4 8 6 0 ] − [ − 6 3 6 − 15 ]
55 A = [ 4 − ( − 6 ) 8 − 3 6 − 6 0 − ( − 15 ) ] = [ 10 5 0 15 ]
Matrix A Now, we can solve for A by dividing both sides by 55:
A = 55 1 [ 10 5 0 15 ] = [ 55 10 55 5 55 0 55 15 ] = [ 11 2 11 1 0 11 3 ]
Solving for B Now that we have A , we can substitute it back into one of the original equations to solve for B . Let's use the second equation:
3 A + 2 B = [ − 2 1 2 − 5 ]
2 B = [ − 2 1 2 − 5 ] − 3 A = [ − 2 1 2 − 5 ] − 3 [ 11 2 11 1 0 11 3 ]
2 B = [ − 2 1 2 − 5 ] − [ 11 6 11 3 0 11 9 ] = [ − 11 22 − 11 6 11 11 − 11 3 11 22 − 0 − 11 55 − 11 9 ] = [ − 11 28 11 8 11 22 − 11 64 ]
Matrix B Finally, we can solve for B by dividing both sides by 2:
B = 2 1 [ − 11 28 11 8 11 22 − 11 64 ] = [ − 11 14 11 4 11 11 − 11 32 ] = [ − 11 14 11 4 1 − 11 32 ]
Final Answer Therefore, the matrices A and B are:
A = [ 11 2 11 1 0 11 3 ]
B = [ − 11 14 11 4 1 − 11 32 ]
Examples
Matrix equations are used in various fields such as computer graphics, physics, and engineering. For example, in computer graphics, transformations such as scaling, rotation, and translation of objects in 3D space can be represented using matrices. Solving systems of matrix equations allows us to determine the parameters of these transformations, enabling us to manipulate and animate objects on the screen. In structural engineering, matrix equations are used to analyze the forces and stresses in complex structures, ensuring their stability and safety. By solving these equations, engineers can optimize the design of buildings and bridges to withstand various loads and environmental conditions.
